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This question is about citric acid, a hydrated tricarboxylic acid - AQA - A-Level Chemistry - Question 8 - 2020 - Paper 2
Question 8
This question is about citric acid, a hydrated tricarboxylic acid. Its formula can be represented as HₓY·zH₂O. 1. A 1.50 g sample of HₓY·zH₂O contains 0.913 g of ox... show full transcript
Worked Solution & Example Answer:This question is about citric acid, a hydrated tricarboxylic acid - AQA - A-Level Chemistry - Question 8 - 2020 - Paper 2
Step 1
1. Show that the empirical formula of citric acid is C₆H₈O₇.
Answer
To determine the empirical formula, we first need to find the number of moles of each element in the sample.
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Calculate the moles of Carbon (C) from the CO₂ produced:
Moles of CO₂ = ( \frac{1.89 \text{ g}}{44.01 \text{ g/mol}} = 0.043 \text{ mol} )
Since each molecule of CO₂ contains 1 C, moles of C = 0.043 mol.
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Calculate the moles of Hydrogen (H) from the H₂O produced:
Moles of H₂O = ( \frac{0.643 \text{ g}}{18.02 \text{ g/mol}} = 0.036 \text{ mol} )
Since each molecule of H₂O contains 2 H, moles of H = 0.036 x 2 = 0.072 mol.
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Calculate the moles of Oxygen (O):
Total mass of sample = 1.50 g. Mass of H and C (calculated) = (0.043 mol × 12.01 g/mol) + (0.072 mol × 1.01 g/mol) = 0.515 g.
Therefore, mass of O = 1.50 g - 0.515 g = 0.985 g.
Moles of O = ( \frac{0.985 \text{ g}}{16.00 \text{ g/mol}} = 0.061 \text{ mol} )
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Now we have the moles: C = 0.043, H = 0.072, O = 0.061.
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Divide by the smallest number of moles (0.043) to get the ratios: C = ( \frac{0.043}{0.043} = 1), H = ( \frac{0.072}{0.043} = 1.67\approx 2), O = ( \frac{0.061}{0.043} = 1.42\approx 1).
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Therefore, the empirical formula in simplest whole number ratio is C₆H₈O₇.
Step 2
2. Show that the value of x = 1.
Answer
We start by calculating the number of moles of the anhydrous sample:
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Moles of anhydrous HₓY:
( M = 210.0 g/mol \rightarrow \text{mass} = 2.74 g \rightarrow \text{moles} = \frac{2.74 g}{210.0 g/mol} = 0.013 , mol. )
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Total mass of the initial sample:
3.00 g. ( M = x + z\times18.02 g/mol = 210.0 .)
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Calculate the value of x using the ratio:
In the hydrated form, we have ( HₓY ext{and z H₂O}). Hence the total moles in the original sample relates to the moles of anhydrous:
( 3.00 g – 2.74 g = 0.26 g \text{ of H₂O} \rightarrow \text{calculate moles of H₂O} )
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Moles of H₂O:
( \frac{0.26 g}{18.02 g/mol} = 0.014 mol \quad \text{Ratio of moles:}\quad \frac{x}{z} = 1:1, \text{thus confirm x = 1.} )
Step 3
Step 4